I decided to spend this summer working through exercises in Hartshorne, and I found myself frustrated by the way I was solving one of them, specifically IV.2.3 on pp. 304-305. No, I'm not asking for a solution to the problem --- I almost have the whole thing solved, as far as I can tell --- what worries me is that I think I'm thinking about this stuff the wrong way.

The problem is about the map from a projective plane curve in characteristic 0 to its dual curve, and it has 8 parts, which ask you to prove things like the fact that bitangents on the original curve correspond to nodes on the dual. I managed to solve all of them except the one about ordinary inflection points on X giving ordinary cusps on X*, but all my solutions involved picking affine charts, finding an ugly formula for the map in question, and computing first and second partial derivatives, and it's all very long and messy and doesn't give any clue as to what's going on until you get to the end and see that the thing you have is 0 or whatever if and only if the condition in the problem is met for some magical reason.

I feel like there must be a more "high-brow" way to approach this object, and the fact that I haven't been able to come up with one seems to speak to the sort of backward way I've been learning about the subject (I just took a class that was very good and covered a lot but was almost completely devoid of examples). There must be some approach to this that actually uses all the machinery that's developed in the rest of the book. Is there a satisfying, pretty way to deal with this thing that I'm missing, something that would tell me why these relationships hold and not just that they hold?

That's what the geometric intuition is for. Move a point along the curve and see what the tangents trace out. And since the duality is an involution, you get to choose at which side you prefer to work. (I don't mean to patronize, I hope it doesn't come across this way.)
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VA.May 11 '10 at 23:48

I can "see the picture" for some of the correspondences -- the node/bitangent picture works pretty well -- but for others I don't see it at all, like cusp/flex. At any rate, what I'd like is some satisfying way to see that that geometric intuition actually works. What I have now feels like magic, and I feel like I just blindly stumbled onto the right answers.
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Nicolas FordMay 12 '10 at 0:44

Check out some of Kleiman's papers from the '80s.
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Felipe VolochMay 12 '10 at 2:08

3 Answers
3

You make your solution feel "fancier", you could start with a more coordinate free approach to the dual curve. For example, try to identify the line bundle on your curve which embeds it into the dual plane. E.g.: let $X \subset \mathbb{P}^2$ be your curve. Then define $X' \subset X \times \mathbb{P}^{2*}$ to be the set of pairs $(x, H)$ where the line $H$ is contained in the tangent space to $X$ at $x$. When $X$ is smooth at least, then this may be identified with the projectivization of the dual of the normal bundle $N_{X/\mathbb{P}^2}$.

Then think about the map $X' \rightarrow X \times \mathbb{P}^{2*} \rightarrow \mathbb{P}^{2*}$. This is the dual curve embedding. The morphism comes from the inclusion of $N^* \subset \Omega_{\mathbb{P}^2}|_X \subset \mathcal{O}(-1)^3$.

At some point, you have to do computations in local coordinates - but setting it up like this may help a little.

Example of local computation:
Locally, the curve looks like $(x(t), y(t), z(t))$. The map to the dual plane is given by $(x(t), y(t), z(t)) \times (x'(t), y'(t), z'(t))$ (cross product! giving the normal to the tangent plane in $\mathbb{C}^3$). If the curve has a simple inflection point, then locally it looks like $(t, t^3, 1)$ and the map to the dual plane is given by $(3t^2, 1, 2t^3)$ - a curve with a simple cusp.

Thank you, I think this is the sort of thing I was looking for. I've been thinking about this problem ever since I posted it, and I still can't quite figure out how to show that an ordinary inflection point on X gives an ordinary cusp on X*, even "the dumb way" in local coordinates, even when X is assumed to be smooth. Do you think this approach would help make that easier? Is there some trick I'm missing?
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Nicolas FordMay 15 '10 at 21:51

Locally at an ordinary inflection point, the curve looks likes y = x^3 + higher order terms. Then show that singularities of the dual curve correspond to the intersection of the curve and the determinant of the Hessian matrix - I'm pretty sure that's how it works.
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mdelandMay 15 '10 at 22:00

This makes sense, but it seems to show just that the ordinary inflection point corresponds to a singularity on the dual, not necessarily an ordinary cusp.
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Nicolas FordMay 15 '10 at 22:15

It has to be a cusp because it has only one branch. The simplest inflection quite naturally gives the simplest cusp, but of course higher order inflection point give higher order cusps!
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quimMay 17 '10 at 8:41

To see that you really do get an ordinary cusp from an ordinary inflection point - I included a local computation above.
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mdelandMay 17 '10 at 17:58

A purely coordinate-free argument might be to think about what happens when you take a curve with a bitangent, and then deform the curve so that the two tangent points come together -- clearly this should produce a flex. We already understand how bitangents give rise to simple nodes on the dual curve. On the side of the dual curve, this deformation would correspond to taking a nodal curve and letting the "nodal part" shrink to a single point. Squint and you get a cusp.

(There used to be also a big computation/handwave in this answer, but on a second reading it was really not illuminating at all.)

See if your library has Gelfand, Kapranov and Zelevinsky's "Discriminants, Resultants, and Multidimensional Determinants", here's the Google Book. The first bit is just on dual varieties in general, and then applies to dual curves, and it's actually fairly accessible, thought the difficulty of the book ramps up rather strongly!

Thanks, this book helped a lot! The one thing I'm still confused about is the flex/cusp correspondence. On p. 20, G, K, and Z say that it "can be easily seen in local coordinates," but I don't quite know what that means. I can follow all their examples, but they're a good deal simpler than the general case.
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Nicolas FordMay 12 '10 at 4:50

Yeah, I didn't have my copy in front of me, so I didn't remember if they did that one specifically. Glad it helped in general, though.
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Charles SiegelMay 12 '10 at 13:56